Graphing Equations Is Useful, V

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Problem

Graphing Equations Is   Useful, V

Solution 1

WLOG, $x\le y\le z\,$ such that $3x+4\le 3y+4\le 3z+4,\,$ implying $z^2\le x^2\le y^2,\,(*)$ from which, say, $z^2-x^2\le 0,\,$ i.e., $(z-x)(z+x)\le 0.\,$ But $z\ge x.\,$ Assume, for now, that $z\gt x.\,$ Then, $x+z\le 0.\,$ We'll consider two cases:

$\mathbf{x\lt z\le 0}$

$x\le y\le z\,\Longrightarrow\,x^2\ge y^2\ge z^2\,\Longrightarrow\,x^2\ge y^2.$

Comparing to (*), $x^2=y^2\,$ and, subsequently, $3y+4=3z+4,\,$ $x=y=z\,$ which contradicts $z\gt x.$

$\mathbf{x\le 0 \le z}\,\text{with }|x|\ge z$

Having $x\le y\le z\,$ suggests two cases: $x\le y\le 0\le z\,$ and $x\le 0\le y\le z.\,$ In the former case, $x^2\ge y^2\,$ which along with (*), would imply $x=y=z=0\,$ which is definitely not a solution. In the latter case, $|x|\ge z,\,$ so that again $x^2\ge y^2,\,$ with the same conclusion.

We may assume now that $z=x\,$ from which $z^2=x^2\,$ and, subsequently, $3x+4=3y+4\,$ such that $x=y\,$ and, therefore, $x=y=z.\,$ So what remains is to solve a quadratic equation $x^2=3x+4.\,$ With two roots, $-1\,$ and $4\,$ we have two solutions: $(x,y,z)=(-1,-1,-1)\,$ and $(x,y,z)=(4,4,4).$

Solution 2

We invert the first equation: $y=\displaystyle \frac{1}{3}(x^2-4)\,$ and consider iterations $x_{k+1}=f(x_k),\,$ with $f(x)=\displaystyle \frac{1}{3}(x^2-4).$

Relevant to the iterations are the points where the graph $y=f(x)\,$ intersects the diagonal, the so called the stationary (or fixed) points of function $f(x).\,$. These are the attractor $(-1,-1)\,$ and the repeller $(4,4).$ Iterations diverge if started with $x_0\gt 4\,$ or $x_0\lt -4.\,$ They converge to $(-1,-1)\,$ for $x_0\in (-4,4).$ There could be a danger that they settle into a finite loop around the attractor. The case of a $3\text{-loop}\,$ would solve our system. To see that there is no loop of any size, we'll prove that the distance to $x=-1\,$ decreases with iterations: $|x_{k+1}+1|\lt |x_k+1|.\,$ In other words, $\displaystyle \left|\frac{1}{3}(x^2-4)+1\right|\lt |x+1|.\;$ I.e., we wish to prove that, for $x\in (-4,4),$

$|x^2-1|\lt |3x+3|.$

Having to deal with absolute values, we'll consider four cases:

$\mathbf{x\ge 1}$

$3x+3\gt x^2-1\,\Longleftrightarrow\,-1\lt x\lt 4,\,$ well covering the range $4\gt x\gt 1.\,$ The required inequality verifies in this case.

$\mathbf{|x|\lt 1}$

$3x+3\gt 1-x^2\,\Longleftrightarrow\,-1\gt x\,\text{or}\,x\lt -2,\,$ well covering the range $|x|\lt 1.\,$ The required inequality verifies in this case.

$\mathbf{-2\le x\lt -1}$

$-3x-3\gt x^2-1\,\Longleftrightarrow\,-2\le x\le -1,\,$ just as needed. The required inequality verifies in this case.

$\mathbf{-4\lt x\le -2}$

In the case, the inequality $-3x-3\gt x^2-1\,$ does not hold, as we just saw. This is because starting in $(-4,-2)\,$ the iterations jump over the $y\text{-axis},\,$ away from $x=-1.\,$ However, in this case, the iterations fall into the interval $(0,4)\,$ from which they continue to converge (according to the previous three cases) to $x=-1.\,$ To see that, let

$-4\lt x\lt -2\,\Longleftrightarrow\, 16\gt x^2\gt 4\,\Longleftrightarrow\,4\gt\displaystyle\frac{1}{3}(x^2-4)\gt 0,$

which completes the proof. The only solutions a finite sequence of iterations may have are defined by the stationary points. In case of three equations, there are two solutions $(x,y,z)=(-1,-1,-1)\,$ and $(x,y,z)=(4,4,4).\,$ Note that the logic does not discriminate between the number of equations. E.g., the system

$\begin{cases} u^2=3v+4&\\ v^2=3w+4&\\ w^2=3x+4&\\ x^2=3y+4&\\ y^2=3z+4&\\ z^2=3u+4& \end{cases}$

has two solutions $(-1,-1,-1,-1,-1,-1)\,$ and $(4,4,4,4,4,4).$

Solution 3 (Illustrated)

Graphing Equations Is   Useful, V. Proof 3, #1

Graphing Equations Is   Useful, V. Proof 3, #2

Clearer.

Graphing Equations Is   Useful, V. Proof 3, #3

Solution 4

Noting that a homogeneous set of equations has a trivial solution $(0,0,0),\,$ let us homogenize the equations. Let $x=p+k,\,$ $y=q+k,\,$ and $z=r+k.\,$ Thus, $x^2=3y+4$ implies $p^2+2pk+k^2=3q+3k+4.\,$ Let $k^2=3k+4\,$ to get rid of the constant terms. There are two solutions $k=-1\,$ and $k=4.\,$

Choose $k=-1.\,$ Thus, $x=p-1,\,$ $y=q-1,\,$ and $z=r-1.\,$ The homogeneous equations become $p^2=2p+3q,\,$ $q^2=2q+3r,\,$ and $r^2=2r+3p.\,$ The trivial solution of $(p=0,q=0,r=0)\,$ gives $(x=-1,y=-1,z=-1).\,$

Alternatively, if we choose $k=4\,$ we get, $x=p+4,\,$ $y=q+4,\,$ and $z=r+4.\,$ The equations become $p^2=-8p+3q,\,$ $q^2=-8q+3r,\,$ and $r^2=-8r+3p.\,$ The trivial solution gives $(x=4,y=4,z=4).\,$

We now need to see if there are any other solutions. To do this, we let $l=x-3/2,\,$ $m=y-3/2,\,$ $n=z-3/2$ $(3/2\,$ is the midpoint of the two solutions already found). The equations become $l^2=-3(l-m)+25/4,\,$ $m^2=-3(m-n)+25/4,\,$ $n^2=-3(n-l)+25/4.\,$ Without loss in generality, let $l\ge mge n.\,$ Thus, the first equation gives $l^2\le 25/4.\,$ The third equation gives $n^2\ge 25/4.\,$ Thus, $l^2\ge 25/4$ as $l\ge n.\,$ This is only possible if $l^2=25/4\,$ and $n^2=25/4.\,$ Adding the three equations, $l^2+m^2+n^2=75/4.\,$ Thus, $m^2=25/4.\,$ Replacing the square terms in the equations with $25/4,\,$ we see that $l=m=n.\,$ The only solutions that satisfy these conditions are the two solutions that we have found.

Acknowledgment

I've lifted this problem from the Imad Zak facebook group. Solution 1 by Imad Zak. The problem fits into a general framework as emerges in Solution 2; Solution 3 is by N. N. Taleb; Solution 4 is by Amit Itagi.

 

Systems of Iterated Equations

  1. Iterations on Monotone Functions
  2. Graphing Equations Is Useful
  3. Graphing Equations Is Useful, II
  4. Graphing Equations Is Useful, III
  5. Graphing Equations Is Useful, IV
  6. Graphing Equations Is Useful, V
  7. Tangential Chaos
  8. Equation in Radicals as a System of Equations
  9. Two Conditions for a Triangle to Be Equilateral

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